a1=1×(2×1+1)=3
當(dāng)n≥2時(shí)
a1+2a2+3a3+…+(n-1)a(n-1)+nan=n{2n+1}
a1+2a2+3a3+…+(n-1)a(n-1)=(n-1)[2(n-1)+1]
兩式相減：
nan=n(2n+1)-(n-1)[2(n-1)+1]
=(2n^2+n)-(n-1)(2n-1)
=(2n^2+n)-(2n^2-3n+1)
=4n-1
an=(4n-1)/n=4-1/n
當(dāng)n=1時(shí),4-1/1=3
a1適合an=4-1/n
所以{an}的通項(xiàng)公式是：
an=4-1/n